This problem is a classic example of integrating concepts from different conic sections: parabolas and circles. We'll determine the tangent lines to the parabola and then use their tangency property with the given circle to find the required value.

---

1. Understanding the Parabola and Deriving its General Tangent Equation

Key Concept: The standard form of a parabola opening left or right is $y^2 = 4ax$. For such a parabola, the equation of a tangent line with a given slope $m$ is $y = mx + \frac{a}{m}$.

Step-by-step Derivation: Our given parabola equation is $2y^2 = -x$. To utilize the standard tangent formula, we must first convert this equation into the standard form $y^2 = 4ax$. Dividing by 2, we get: $y^2 = -\frac{1}{2}x$ Now, we compare this with the standard form $y^2 = 4ax$: $4a = -\frac{1}{2}$ Solving for $a$, we find: $a = -\frac{1}{8}$ This negative value of $a$ indicates that the parabola opens to the left.

Next, we substitute this value of $a$ into the general tangent equation $y = mx + \frac{a}{m}$: $y = mx + \frac{-1/8}{m}$ $y = mx - \frac{1}{8m}$ This equation represents any tangent line to the parabola $2y^2 = -x$ in terms of its slope $m$.

Why this step is taken: We need to find the specific tangent lines ($l_1$ and $l_2$). By first establishing the general form of a tangent specific to our parabola, we create a framework to determine the slopes of these particular tangents in the next step. Without this, we wouldn't have a starting point to link the lines to the parabola's properties.

Tip: Always be meticulous with the sign of $a$. A positive $a$ means the parabola opens right, while a negative $a$ means it opens left. Similarly, for $x^2=4ay$, a positive $a$ means it opens upwards, and a negative $a$ means it opens downwards. Using the correct standard form and tangent equation is crucial.

---

2. Finding the Slopes of the Tangent Lines ($l_1$ and $l_2$)

Key Concept: If a line passes through a specific point, the coordinates of that point must satisfy the line's equation.

Step-by-step Calculation: We are given that the tangent lines $l_1$ and $l_2$ are drawn from the point $(2,0)$ to the parabola. This implies that the point $(2,0)$ lies on both these tangent lines. We substitute the coordinates of the point $(2,0)$ (i.e., $x=2$ and $y=0$) into our general tangent equation $y = mx - \frac{1}{8m}$: $0 = m(2) - \frac{1}{8m}$ Now, we solve this equation for $m$ to find the slopes of the specific tangent lines. $0 = 2m - \frac{1}{8m}$ To eliminate the fraction and simplify, we multiply the entire equation by $8m$. We can safely assume $m \neq 0$, because if $m=0$, the tangent equation would be $y=0$. The line $y=0$ (the x-axis) is tangent to the parabola $2y^2=-x$ at the origin $(0,0)$, but it does not pass through the point $(2,0)$. $0 \cdot (8m) = (2m) \cdot (8m) - \left(\frac{1}{8m}\right) \cdot (8m)$ $0 = 16m^2 - 1$ Rearranging the terms to solve for $m^2$: $16m^2 = 1$ $m^2 = \frac{1}{16}$ Taking the square root of both sides gives us two possible values for $m$: $m = \pm \sqrt{\frac{1}{16}}$ $m = \pm \frac{1}{4}$ These two values, $m_1 = \frac{1}{4}$ and $m_2 = -\frac{1}{4}$, are the slopes of the two tangent lines $l_1$ and $l_2$.

Why this step is taken: The point $(2,0)$ is an external point to the parabola. From an external point, two tangents can generally be drawn to a parabola. By substituting the coordinates of this point into the general tangent equation, we impose the condition that the tangent must pass through $(2,0)$, thereby uniquely determining the slopes of those two specific tangents.

Common Mistake: A frequent error is to forget the $\pm$ when taking the square root of $m^2$. This would lead to finding only one tangent line instead of the two expected from an external point, which would then lead to an incomplete solution. Always remember that a quadratic equation typically yields two solutions.

---

3. Formulating the Equations of Tangent Lines $l_1$ and $l_2$

Key Concept: Once the slope $m$ is known, we can find the specific equation of the tangent line by substituting $m$ back into the general tangent equation. For subsequent calculations involving distance, it's often most convenient to express lines in the general form $Ax+By+C=0$.

Step-by-step Calculation: Now we use the two values of $m$ we found and substitute them back into the general tangent equation $y = mx - \frac{1}{8m}$ to obtain the specific equations of $l_1$ and $l_2$.

• For $m = \frac{1}{4}$ (let's call this $l_1$): $y = \left(\frac{1}{4}\right)x - \frac{1}{8(1/4)}$ $y = \frac{1}{4}x - \frac{1}{2}$ To convert this to the general form $Ax+By+C=0$, we multiply the entire equation by 4 to clear the denominators: $4y = x - 2$ Rearranging the terms: $x - 4y - 2 = 0$ This is the equation of line $l_1$.

• For $m = -\frac{1}{4}$ (let's call this $l_2$): $y = \left(-\frac{1}{4}\right)x - \frac{1}{8(-1/4)}$ $y = -\frac{1}{4}x - \left(-\frac{1}{2}\right)$ $y = -\frac{1}{4}x + \frac{1}{2}$ Again, multiply by 4 to get the general form: $4y = -x + 2$ Rearranging the terms: $x + 4y - 2 = 0$ This is the equation of line $l_2$.

So, the two tangent lines are $l_1: x - 4y - 2 = 0$ and $l_2: x + 4y - 2 = 0$.

Why this step is taken: We need the explicit equations of $l_1$ and $l_2$ in the general form ($Ax+By+C=0$) because this specific format is required to use the formula for the perpendicular distance from a point to a line, which is a crucial part of the tangency condition for the circle in the next steps.

---

4. Analyzing the Circle and Its Tangency Condition

Key Concepts:

1. The standard equation of a circle is $(x-h)^2 + (y-k)^2 = R^2$, where $(h,k)$ is the center and $R$ is the radius.
2. Condition for Tangency: A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is exactly equal to the radius of the circle.
3. Distance from a Point to a Line: The perpendicular distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Step-by-step Identification: The given equation of the circle is $(x-5)^2 + y^2 = r$. Comparing this with the standard form $(x-h)^2 + (y-k)^2 = R^2$:

• The center of the circle $C = (h,k) = (5,0)$.
• The square of the radius is $R^2 = r$, so the radius $R = \sqrt{r}$.

Why this step is taken: To apply the tangency condition effectively, we first need to extract the fundamental properties of the circle: its center and its radius. This information, along with the equations of the tangent lines we just found, will allow us to set up an equation to solve for the unknown value $r$.

---

5. Applying the Tangency Condition to Find the Value of $r$

Key Concept: As established, for a line to be tangent to a circle, the perpendicular distance from the circle's center to that line must be equal to the circle's radius.

Step-by-step Calculation: We will calculate the perpendicular distance from the center of the circle $(x_0, y_0) = (5,0)$ to one of the tangent lines. Let's choose $l_1: x - 4y - 2 = 0$. For this line, we have $A=1$, $B=-4$, and $C=-2$. Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$: $d = \frac{|(1)(5) + (-4)(0) + (-2)|}{\sqrt{(1)^2 + (-4)^2}}$ $d = \frac{|5 - 0 - 2|}{\sqrt{1 + 16}}$ $d = \frac{|3|}{\sqrt{17}}$ $d = \frac{3}{\sqrt{17}}$ Since $l_1$ is tangent to the circle, this distance $d$ must be equal to the radius of the circle, $R = \sqrt{r}$: $\sqrt{r} = \frac{3}{\sqrt{17}}$ (Note: If we had chosen line $l_2: x + 4y - 2 = 0$, the distance calculation would be $d = \frac{|(1)(5) + (4)(0) + (-2)|}{\sqrt{(1)^2 + (4)^2}} = \frac{|3|}{\sqrt{17}}$, yielding the exact same result. This consistency is expected due to the symmetry of the problem, where the lines are $x \pm 4y - 2 = 0$ and the center of the circle is on the x-axis.)

Now, to find $r$, we square both sides of the equation: $(\sqrt{r})^2 = \left(\frac{3}{\sqrt{17}}\right)^2$ $r = \frac{3^2}{(\sqrt{17})^2}$ $r = \frac{9}{17}$

Why this step is taken: This is the pivotal step where we connect the information about the parabola's tangents to the circle. By applying the geometric tangency condition (distance from center to line equals radius), we establish an equation that allows us to directly solve for the unknown radius $r$. It's the critical bridge between the two conic sections.

Common Mistake: Forgetting the absolute value in the numerator of the distance formula can lead to incorrect signs, although in this particular case, $|3|$ is just $3$. Also, ensure to square both the numerator and denominator correctly when solving for $r$.

---

6. Calculating the Final Required Value

Step-by-step Calculation: The question asks for the value of $17r$. We have found $r = \frac{9}{17}$. Substitute this value into the expression $17r$: $17r = 17 \times \left(\frac{9}{17}\right)$ $17r = 9$

Why this step is taken: This is the final step to provide the answer in the specific format requested by the problem statement.

---

Summary and Key Takeaways

The final value of $17r$ is $\boxed{9}$.

This problem is an excellent illustration of how different concepts from coordinate geometry, specifically involving parabolas and circles, can be combined. The key steps and takeaways include:

1. Standardization: Always convert given conic section equations into their standard forms ($y^2=4ax$, $(x-h)^2+(y-k)^2=R^2$) to easily extract parameters like $a$, center $(h,k)$, and radius $R$.
2. General to Specific: Use general tangent equations (e.g., $y=mx+a/m$) and the coordinates of external points to find the specific equations of the tangent lines. This involves solving for the slopes $m$.